Is there a quick way to arrive to this partial fraction decomposition?

Question

I'm reading a book where the author claims without showing the work that after we go through with the algebra of partial fractions we arrive to the formula (note $|z|<1$): $$\frac{1}{(1-z)(1-z^2)(1-z^3)} = \frac{1}{6}\frac{1}{(1-z)^3}+\frac{1}{4}\frac{1}{(1-z)^2}+\frac{1}{4}\frac{1}{(1-z^2)}+\frac{1}{3}\frac{1}{(1-z^3)}$$

Naturally, I'm trying to reproduce the result, but I'm taking a very naive approach, namely I've expressed the LHS as a product of irreducible factors over $\mathbb{R}$ and I am trying to determine the coefficients: $$\frac{1}{(1-z)^3}\frac{1}{(1+z)}\frac{1}{(z^2+z+1)} = \frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-z)^3}+\frac{D}{(1+z)}+\frac{Ex+F}{(z^2+z+1)}$$

This seems to have some drawbacks though, because on top of being long winded, once I do obtain all the coefficients I will have to recombine some of the terms to arrive at the author's answer. My question is then: is there some sort of trick which we can use here which I am not aware of, or do we have to suffer through the algebra patiently?

Answer 1

First of all observe that

$$f(z) = \frac{-1}{(z-1)(z^2 - 1)(z^3 - 1)} = \frac{-1}{(z-1)^3(z+1)(z+1/2-i\sqrt{3}/2)(z+1/2+i\sqrt{3}/2)}$$

Remembering now the cauchy integral formula

$$ g^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_{\gamma} \frac{g(z)}{(z-z_0)^{n+1}}dz $$

If, for example, you consider

$$ g(z) = \frac{-1}{(z-1)^3(z+1/2-i\sqrt{3}/2)(z+1/2+i\sqrt{3}/2)} $$

we have

$$ g(-1) = \frac{1}{2\pi i} \oint_{\gamma} \frac{g(z)}{z+1}dz = \frac{-1}{(-8)(-1/2-i\sqrt{3}/2)(-1/2+i\sqrt{3}/2)} $$

which is the coefficient of $A/(z+1)$, you can iterate through all the poles to obtain all the other coefficients, the only "tricky" one is the pole "z=1", which is a third order pole, so you it would a the calculation of a second derivative.

Answer 2

Here's a bit of a shortcut (if you already know the coefficients).

Multiply the whole thing by $(1-z)(1-z^2)(1-z^3)$, to get \begin{align} & \frac{1}{6}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z)^3}+\frac{1}{4}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z)^2}+\frac{1}{4}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z^2)}+\frac{1}{3}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z^3)}\\ \ \\ &= \frac{(1+z)(1+z+z^2)}{6}+\frac{(1+z)(1-z^3)}{4}+\frac{(1-z)(1-z^3)}{4}+\frac{(1-z)(1-z^2)}{3}\\ \ \\ &= \frac{2(1+z)(1+z+z^2)+3(1+z)(1-z^3)+3(1-z)(1-z^3)+4(1-z)(1-z^2)}{12}\\ \ \\ &=\frac{2(1+z)(1+z+z^2)+6(1-z^3)+4(1-z)(1-z^2)}{12}\\ \ \\ &=\frac{2z^3+4z^2+4z+2+6-6z^3+4-4z-4z^2+4z^3}{12}\\ \ \\ &=\frac{12}{12}=1. \end{align}